Physical interpretation of the definition of angular momentum in classical mechanics

Question

To what I understand, the following is a valid way to introduce the angular momentum $\mathbf L$ in the Lagrangian system of a rigid body. We can consider the extended configuration space to be $M\times \mathbb{R}=SO(3)\times \mathbb R$. For a left invariant Riemannian metric $g$ in $SO(3)$ we can define the Lagrangian $\mathcal L:TM\to \mathbb R$ as the map $v\mapsto \frac 1 2 g(v,v)$. Suppose $v\in TM$, we can define the map $\Omega$ from TM into the Lie algebra $\mathfrak g=so(3)=T_eM$ of $SO(3)$ by $$v\mapsto \Omega_v=\big(L_{g^{-1}}\big)_*v\quad \text{for }v\in T_gM,$$ where $L_g$ is the left translation on $M$. Hence, the Lagrangian take the form $\mathcal L(v)=g(\Omega_v,\Omega_v)$. $\Omega_v$ is called the angular velocity. Furthermore, since the killing form $B:\mathfrak g\times \mathfrak g\to \mathbb R$ is a bilinear non-degenerate form on $so(3)$ (since it is a semisimple Lie algebra) defined by $B(\alpha,\beta)=-1/2 \; \text{Tr } (\alpha\beta)$. Then there exists a symmetric endomorphism $\mathbf L:\mathfrak g\to \mathfrak g$ with the property that $$\mathcal L(v)=B(\mathbf L\Omega_v,\Omega_v).$$ Such map (for fixed $v\in TM$ and thus fixed $\Omega_v$) is called the Angular momentum of the body. This is a nice and elegant way to define the Angular momentum, for the eigenvalues of $\mathbf L$ are the principal axes of inertia, and its corresponding eigenvalues, the principal moments of inertia. But what is still bothering me, is that the only physical intuition I get from such definition is the fact that a rigid body, can ultimately be characterized by translations and rotations (here I assume, translations are not being considered). But the definitions of the Lagrangian and of the angular momentum; seem to be too artificial to me. Is there a nice way to see this?